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Is it possible to find all finite group which has a subgroup of certain order $n$ (here say for example $2$) ? I am thinking about this problem for few days, but not getting how to approach.

I am thinking that the theorem -

Every finite group is isomorphic to a subgroup of $S_n$.

can give something about it. Can anyone answer please? I have searched for it, but it is not posted here and not available in internet.

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    $\begingroup$ If I'm not mistaken, the trivial group of order 1 is a subgroup of all groups. Thus, a special case of your problem is equivalent to the classification of all finite groups, which is much harder than the classification of all finite simple groups, which is a problem that took decades of work by hundreds of group theorists. Good luck! $\endgroup$ – Display name Jul 28 '18 at 5:26
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    $\begingroup$ @Displayname Ok, then for some specific $n$ is it possible to find all groups, which has subgroup of order $n$ ? as, I have mentioned suppose - $n=2$. $\endgroup$ – user579462 Jul 28 '18 at 5:34
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    $\begingroup$ Also, every finite group whose order is divisible by a prime number $p$ has a subgroup of order $p$ (Cauchy's theorem). Thus, another special case of your problem is equivalent to the classification of all finite groups that are not $p'$-groups. $\endgroup$ – Orat Jul 28 '18 at 5:42
  • $\begingroup$ Well, to take the help of Calay's theorem, can we find all subgroups of $S_n$ that has subgroup of order $2$ ? If we can find, then it can be stated - "The groups which has subgroup of order $2$ are $\cdots$, upto isomorphism" $\endgroup$ – user579462 Jul 28 '18 at 6:26
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    $\begingroup$ @KenOno By Cauchy's Theorem, every group of even order has a subgroup of order $2$. By Cayley's theorem, every group is isomorphic to a subgroup of $S_n$ for some $n$. So finding all such subgroups is equivalent to finding all groups of even order. But, given any finite group $G$, the group $G \times \mathbb{Z}/2\mathbb{Z}$ is a group of even order, so again, this is equivalent to finding all finite groups, a problem very far from being solved as it stands now.. $\endgroup$ – Henrique Augusto Souza Jul 28 '18 at 6:35
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First, the result regarding $S_n$ that you are referring to is known as "Cayley's theorem".

In fact, a necessary condition is $n$ divides $|G|$, the order of the group $G$.

Indeed, if there exists a subgroup of $G$ that has order $n$, then $n$ will divide $|G|$ by Lagrange's theorem.

Under a certain condition, a converse holds.

Suppose that $n$ divides $|G|$, and pick Sylow groups $P_1, \ldots, P_r$, where we assume that $n = p_1^{k_1} \cdots p_r^{k_r}$ and $P_j$ is a $p_j$-Sylow group.

Now if $P$ is any $p$-group, then it has a normal subgroup of order $p^m$, whenever $p^m$ divides $|P|$. For $m=1$ this is Cauchy's theorem, and for general $m$ use Cauchy's theorem and divide out the corresponding cyclic group and then use induction.

Thus, pick for each $j$ a subgroup $Q_j \le P_j$ of order $p_j^{k_j}$. Then the $Q_j$ are pairwise disjoint for order reasons, so that they define a subgroup $$ H := \langle Q_1, \ldots, Q_r \rangle $$ that is a direct product of $Q_1, \ldots, Q_r$ (if each $Q_j$ is a normal subgroup of $H$) and hence has order $n$.

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    $\begingroup$ That may, unfortunately, be so, since H may permute the subgroups of the Sylow p-groups by conjugation. This happens when H contains an element of order p, say, which is not in the corresponding p-group generator. $\endgroup$ – AlgebraicsAnonymous Jul 28 '18 at 6:19
  • $\begingroup$ Even though in the abelian case this can't happen; this we see from the explicit description of the generated subgroup and the formula for the order of the sum of elements whose gcd is 1. $\endgroup$ – AlgebraicsAnonymous Jul 28 '18 at 6:22
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As noted in the comments by Display name and Orat, it is (as of today) improbable for general or even for a given $n$ to find all groups containing a subgroup of order $n$ in the sense of classification - to formulate a list of groups such as that any group is isomorphic to one in the list.

On the other hand, sometimes necessary and/or sufficient conditions can be given. One simple necessary condition is that $n$ divides the order of $G$, and one simple sufficient condition is that $G$ has order $n$.

For specific $n$, better conditions can be given. If $n = p$ is a prime, then Cauchy's theorem states that a necessary and sufficient condition is that $p$ divides the order of $G$. This is strengthened by Sylow's theorems: a necessary and sufficient condition for a finite group $G$ to have a subgroup of order $p^a$ is that $p^a$ divides the order of $G$. In terms of classification, even $p$-groups are hard to tell, let alone all groups who contain them.

When $n$ is no longer a prime, things get complicated. If we factor $n = p_1^{a_1}\cdots p_k^{a_k}$, then for a group $G$ whose order is a multiple of $n$ we can easily find subgroups $H_i$ with order $p_i^{a_i}$ for every $1 \leq i \leq k$. But we cannot guarantee in general that their product is again a subgroup (altough it will have the correct order as a set and every subgroup of order $n$ can be written in such way). I don't have a construction out of my pocket, but I believe that arbitrarily bad behaved examples can be found.

If your group $G$ is nilpotent, than each of the $H_i$ in the above construction are normal in $G$ and you can guarantee that the product $H = H_1\cdots H_k$ is again a subgroup of $G$ of order $n$. Classify all nilpotent finite groups - that's also a hard problem.

According to GroupWiki, your property lies somewhere between supersolvable (which is weaker than nilpotency) and solvable, as the alternating group on four letters $A_4$ is a solvable group that does not satisty this property. It also states that any group satisfying it must be solvable. Note that I'm talking about an even stronger property: a group $G$ having subgroups of every order possible (at least one subgroup for each divisor of it's order). If we talk about a fixed $n$, then we go back to the original problem: if $H$ has order $n$ and $G$ is any finite group, $G\times H$ is a group containing a subgroup of order $n$, and their classification involves classifying all finite groups.

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  • $\begingroup$ Can you say something about the proof for the nilpotent case? $\endgroup$ – AlgebraicsAnonymous Jul 28 '18 at 6:26
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    $\begingroup$ @AlgebraicsAnonymous A nilpotent group is a direct product of it's Sylow subgroups. So each Sylow subgroup is normal in G, and by the direct product structure, the same is true of all the subgroups of it's Sylow subgroups. See: math.stackexchange.com/questions/24220/… $\endgroup$ – Henrique Augusto Souza Jul 28 '18 at 6:28

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